Evaluating the sum $1\cdot 10^1 + 2\cdot 10^2 + 3\cdot 10^3 + \dots + n\cdot 10^n$ How can I calculate 
$$1\cdot 10^1 + 2\cdot 10^2 + 3\cdot 10^3 + 4\cdot 10^4+\dots  + n\cdot 10^n$$
as a expression, with a proof so I could actually understand it if possible?
 A: Put
$$ S_n = \sum_{k=1}^{n} k\cdot 10^k.$$
Then:
$$ 9 S_n = (10-1)S_n = \sum_{k=1}^{n}k\cdot 10^{k+1}-\sum_{k=1}^{n}k\cdot 10^k=\sum_{k=2}^{n+1}(k-1)\cdot 10^k-\sum_{k=1}^{n}k\cdot 10^k$$
hence:
$$ 9 S_n = n\cdot 10^n -\sum_{k=1}^n 10^k = n\cdot 10^n-\frac{10^{n+1}-10}{9},$$
so:
$$ S_n = \color{red}{\frac{10}{81}\left(1+(9n-1)\cdot 10^n\right)}.$$
A: Your expression is 
$$S = \sum_{k=1}^n k 10^k$$
We can pull out a factor of $10$ to get
$$S = 10 \sum_{k=1}^n k 10^{k-1}$$
Now, consider the function
$$f(x) = \sum_{k=1}^{n}kx^{k-1}$$
So $S = 10f(10)$. For $x \neq 1$, we can rewrite the sum as follows:
$$f(x) = \begin{align}
\frac{d}{dx}\sum_{k=1}^{n} x^k
&= \frac{d}{dx} \frac{x - x^{n+1}}{1-x}
\end{align}$$
where we have used the fact that $(x + x^2 + \cdots + x^n)(1-x) = x - x^{n+1}$ due to telescoping. 
Therefore the desired sum is
$$S = 10f(10) = 10 \left.\left(\frac{d}{dx} \frac{x - x^{n+1}}{1-x}\right)\right|_{x=10}$$
I'll let you carry out the differentiation to finish the problem.
A: Let $$S_n = 1*10^1+2*10^2+3*10^3 + \cdots n*10^n$$
$$10S_n = 1*10^2+2*10^3+3*10^4  + \cdots n*10^{n+1}$$
$$S_n - 10S_n = 1*10^1 + 1*10^2 + 1*10^3 +\cdots1*10^n-n*10^{n+1}$$
$$-9S_n = 10*\frac{1-10^n}{1-10} - n*10^{n+1}$$
Simplifying, $$S = \frac{10}{81}*(9*10^n n-10^n+1)$$
